Proposition 33

Table of Contents

PROPOSITION 33. Problem.

If a body is always attracted to the centre of force C (Fig. 38), to determine the curve AM, upon which the body is moving with a uniform angular motion around the centre C.

Solution

Let A be the highest point of the curve, where the curve is normal to the radius AC, and let the speed of the body at A correspond to b and AC = a; the angular speed at A = ab , to which quantity the angular speed at some particular point M must be equal. Putting CM = x, to which CP is taken equal, and the centripetal force at M is equal to P; then the speed at M corresponds to the height b − Pdx ,

…

with the integration of Pdx thus taken, so that it vanishes on putting x = a. With the tangent MT drawn, the perpendicular sent from C to that is called CT = p; then we have x : p = Mm : mn. [p. 137; again, the infinitesimal triangle Mnm and the finite triangle MTC are similar] On account of this, the speed passing along mn is equal to and the angular speed is equal to which must be equal to ab . Hence the following equation is produced : or

…

With centre C and with radius BC = 1 the arc of the circle BS is described, which is called equal to s, then 1 : ds = x : mn , and hence mn = xds and Mm = ( dx 2 + x 2ds 2 ) . Since now we have x : p = ( dx 2 + x 2ds 2 ) : xds , the equation becomes

With which value substituted in the equation found gives .. and hence ..

From which equation the curve sought can be constructed. Q.E.I.

Corollary 1

When x is made smaller, then the larger b − Pdx becomes, whereby when x is .. made smaller, with that also x becomes less, or the sine of the angle CMT.

For …

Corollary 2

Again, as by hypothesis since in this equation x cannot become greater than a ; for it would make p > x. On account of which none of the radii CM can be normal to the curve, unless it is at a maximum, clearly equal to AC.

Scholium 1

For whatever hypothesis of centripetal force, it is satisfactory for a circle with centre C to be described [p. 138] ; for the body must be moving uniformly on a given circle. Moreover even if the general equation does not seem to include the circle, yet no less it must be contained, as we have now intimated above.

[Part of Euler’s problem was the imperfect state of affairs at the time regarding dynamics : we would now make some reference to angular momentum, which he was later to clarify, in the analysis. This of course makes these special curves all the more fascinating for us to look back on; thus, Euler’s use of a potential energy or a work related function was introduced as a means of simplifying problems – only to be discovered later that this was how the world really worked, although Euler understood that he had to base his dynamics on known physical facts, such as Huygens’ pendulum as a means of finding the acceleration of gravity, and Galileo’s inclined plane : it was the latter that gave rise to the potential energy function that relates height fallen from rest to the square of the final speed.]

Scholium 2

No other curve going around the centre except the circle is able to satisfy the condition sought. For in curves of this kind it is not possible that all the lines drawn from the centre and normal to the curve are equal to each other.

Therefore which curves besides the circle which solve the problem, these must pass through the centre C itself, so that not more than one radius MC is normal to the curve. These curves are of this kind that we look at in the following example.

Example

Let the centripetal force be directly proportional to the distances from the centre or P = xf ; the equation becomes : …

With which substituted, the following equation is produced for the curve :

Now the arc is .. ∫ ( adx− x ) , the sine of which is ax with the total sine arising .. equal to 1. This arc is denoted by A. ax . Let the sine of the arc BS = t, then s = A.t; hence the equation becomes : …

Or the arc, of which the cosine is ax , is equal to :

Hence the construction of the curve easily follows on, and it is an algebraic curve, whenever … is a rational number.

Let .. giving the integral of this, by means of imaginary logarithms, is given by …

…

The perpendicular MQ = y is sent from M to AC and on putting CQ = u there is .. 1 : t = x : y and so t = x .

Hence, …

Or, if m = 2 or bf = a6 , then this equation is obtained : .. which reduced gives : ..

or …

But if an equation is desired between the orthogonal coordinates u and y, then it is this equation of the sixth order :

In this curve the applied line is at a maximum if x = 2b .. or if we take

for then it becomes:

In other examples of this, with the values m , the maximum applied line is where …